in N1." The ratio AY1 / AN1 is equal to the slope of the
line which connects the mean yield values for 0 and for
150 kg/ha of nitrogen. This is also the average slope of
the curve between 0 and 150 kg/ha of nitrogen or
equivalent to 75 kg of nitrogen (Fig. VI-4). Hence, the
value of AY1 / AN1 = 8.83 is plotted on the lower portion
of Fig. VI-4 for N = 75. The slope of the response
surface between 150 and 300 kg/ha of nitrogen is
calculated the same way. Here, AY2 is 300 and AN2 is 150,
so AY2 / AN2 = 2. This value is plotted midway between
150 and 300 or equivalent to N = 225. The straight line
drawn between the two values of AY/AN plotted on the lower
part of Fig. VI-4 is the function of the slope, or the
first derivative of the quadratic response surface.
 Having two points on a straight line, it is simple to
calculate the equation of this line. The slope is derived
in the same manner as the slope for the curve. Find the
change in value of the derivative and divide that by the
change in value of N. In this case there are two known
points. For N = 75, the value of AY/AN is 8.83 and for N
= 225, the value of AY/AN is 2. Hence, for the straight
line the value of A(AY/AN) = 8.83 2.00 = 6.83. The
difference between 225 and 75 is 150, so the slope of the
straight line is: A(AY/AN)/AN = -6.83 / 150 = -0.0455. It
is negative because the value AY/AN diminishes between the
values of N = 75 and N = 225.
 To this point, then, a portion of the equation of the
straight line is known. That is, the value of the slope b
equals -0.0455, so Y = a 0.0455 N. To find the value of
a, known values of AY/AN and N can be substituted to solve
the equation for a. For N = 75, the value of AY/ AN =
8.83. Substituting these in the above equation, produces
-8.83 = a 0.0455 (75). Solving this equation for a
gives a = 12.25. The equation of the derivative of the
quadratic curve, then, is

 AY/AN = 12.25 0.0455 N

 Mathematically, the quadratic response curve is the
integral of the straight line equation above. A simple
means of finding the integral of a straight line is to
multiply the equation by N (in this case) and divide the
slope of the straight line by 2. The result,